1
JEE Main 2021 (Online) 27th July Morning Shift
+4
-1

A capacitor of capacitance C = 1 $$\mu$$F is suddenly connected to a battery of 100 volt through a resistance R = 100 $$\Omega$$. The time taken for the capacitor to be charged to get 50 V is :

[Take ln 2 = 0.69]
A
1.44 $$\times$$ 10$$-$$4 s
B
3.33 $$\times$$ 10$$-$$4 s
C
0.69 $$\times$$ 10$$-$$4 s
D
0.30 $$\times$$ 10$$-$$4 s
2
JEE Main 2021 (Online) 27th July Morning Shift
+4
-1
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be :

(Given area of plate = A)

A
$${{15} \over {34}}{{K{\varepsilon _0}A} \over d}$$
B
$${{15} \over 6}{{K{\varepsilon _0}A} \over d}$$
C
$${{25} \over 6}{{K{\varepsilon _0}A} \over d}$$
D
$${9 \over 6}{{K{\varepsilon _0}A} \over d}$$
3
JEE Main 2021 (Online) 27th July Morning Shift
+4
-1
Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :
A
$${V \over {K + 2}}$$
B
$${V \over K}$$
C
$${{3V} \over {K + 2}}$$
D
$${{3V} \over K}$$
4
JEE Main 2021 (Online) 25th July Evening Shift
+4
-1
If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :
A
$${q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)$$
B
$${q_b} = {q_f}\left( {1 - {1 \over k}} \right)$$
C
$${q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)$$
D
$${q_b} = {q_f}\left( {1 + {1 \over k}} \right)$$
EXAM MAP
Medical
NEET